Question #c0369

1 Answer
Nov 30, 2015

Answer:

#["H"_3"O"^(+)] = ["OH"^(-)] = 2.35 * 10^(-7)"M"#

Explanation:

The balanced chemical equation for water's self-ionization reaction looks like this

#2"H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-)#

The equilibrium constant for this reaction would look like this

#K_(eq) = ( ["H"_3"O"^(+)] * ["OH"^(-)])/(["H"_2"O"]^2)#

Now, when dealing with dilute aqueous solutions, the concentration of water can be thought of as being constant. This means that you can write

#overbrace(K_(eq) * ["H"_2"O"]^(2))^(color(blue)(K_W)) = ["H"_3"O"^(+)] * ["OH"^(-)]#

This expression, #K_(eq) * ["H"_2"O"]^2#, is called water's self-ionization constant, or water's ion product constant.

Now, you know that this constant is equal to #5.5 * 10^(-14)# at #50^@"C"#. Since the self-ionization reaction produces equal number of moles of hydronium and hydroxide ions, you can say that #x# represents the molarity of these species in aqueous solution.

This means that you can write

#K_W = x * x = x^2#

In your case, you would have

#5.5 * 10^(-14) = x^2 implies x= sqrt(5.5 * 10^(-14)) = 2.35 * 10^(-7)#

Therefore,

#["H"_3"O"^(+)] = ["OH"^(-)] = color(green)(2.35 * 10^(-7)"M")#

As an interesting comment, the pH of water at this temperature would be

#"pH" = - log( ["H"_3"O"^(+)])#

#"pH" = - log( 2.35 * 10^(-7)) = 6.63#

The important thing to realize here is that the water would still be neutral because you have equal concentrations of hydronium and hydroxide ions.